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LHO - coupled oscillators

  1. Jan 17, 2007 #1
    1. The problem statement, all variables and given/known data

    The question then goes on to say:
    Decompose the resulting oscillation as a superposition of symmetric and antisymmetric mode oscillations. Hence give A and B in terms of C

    2. Relevant equations

    3. The attempt at a solution
    Well as of yet I'm not sure I fully understand the question. I think all its asking me to do is to combine the two waves to give one equation that describes the resultant wave.

    As I understand the theory behind it, [tex]\omega_A[/tex] is 180 degrees out of phase with [tex]\omega_S[/tex]. Does this mean that the superposition of the waves cancels each other out? Or does the oscillation of the coupling spring contribute to this?

    As far as I can gleam as well, A = C and B = -C. Is this right?
  2. jcsd
  3. Jan 17, 2007 #2


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    In the general case, the phases of the two waves are completely independent. The general solution is

    x1s = Acos(ws t + ps) = x2s
    x1a = Bcos (wa t + pa) = -x2a

    where ps and pa are two independent phase angles. In the question, they have given you the particular solution that matches the initial conditions they specified, and that has ps = pa = 0.

    Your "A = C and B = -C" is not quite right. At time t = 0 you have

    x1 = x1s + x1a = A + B = C and
    x2 = x2s + x2a = A - B = 0.

    The waves don't cancel out because they have different frequencies. The formulas given in the question show that the coupling spring affects the frequency of the antisymmetric mode, but not the symmetric one. That makes sense - in the symmetric mode the coupling spring does not change length so its stiffness is irrelevant, but in the antisymmetric mode it does change length and its stiffness affects the frequency of the mode.
    Last edited: Jan 17, 2007
  4. Jan 18, 2007 #3

    I'm not meant to write it as one function like Asinx + Bcosx = Xcos(x+w)?

    I'm still not 100% sure what's going on with this question and what its asking me.

    Is the x1 = x1s + x1a = A + B (and simililar for x2) part the first part of the question? Then that is used to give A and B in terms of C?
  5. Jan 18, 2007 #4


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    As far as I can see, all they want you to do is write the motion as

    x1 = A cos w1 t + B cos w2 t
    x2 = A cos w1 t - B cos w2 t

    and use the conditions at t = 0 to find A and B in terms of C.

    I agree it seems a strange question. They seem to have done all the hard stuff for you.
  6. Jan 18, 2007 #5
    So I've done that and gotten that A=B=C/2.

    Does that sound about right?
  7. Jan 18, 2007 #6


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    Yep :approve:
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